Differential Equations Quiz: Coupled Mass Spring Systems

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Coupled Mass Spring SystemsQuestion 1 of 5

A coupled mass-spring system consists of two masses $m_1 = 2$ kg and $m_2 = 1$ kg connected by springs with constants $k_1 = 8$ N/m, $k_2 = 4$ N/m, and $k_3 = 2$ N/m, where $k_1$ connects $m_1$ to a fixed wall, $k_2$ connects the two masses, and $k_3$ connects $m_2$ to a fixed wall. If the system is set into motion with initial conditions $x_1(0) = 1$ , $x_2(0) = 0$ , $x_1'(0) = 0$ , $x_2'(0) = 2$ , what is the coefficient matrix for the system of differential equations $\mathbf{x}'' = A\mathbf{x}$ ?

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Differential Equations Quiz

Differential Equations Quiz: Coupled Mass Spring Systems

Practice Coupled Mass Spring Systems in Differential Equations with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Coupled Mass Spring Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Differential Equations.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

(correct answer)

Explanation: For the coupled system, we have

m_1 x_1'' = -k_1 x_1 + k_2(x_2 - x_1)

and

m_2 x_2'' = -k_2(x_2 - x_1) - k_3 x_2

. This gives us

2x_1'' = -8x_1 + 4(x_2 - x_1) = -12x_1 + 4x_2

and

x_2'' = -4(x_2 - x_1) - 2x_2 = 4x_1 - 6x_2

. Therefore

x_1'' = -6x_1 + 2x_2

and

x_2'' = 4x_1 - 6x_2

, giving matrix A as shown. Choice B incorrectly calculates the diagonal entries. Choice C incorrectly assumes symmetric coupling. Choice D fails to divide by the masses properly.

Question 2

A coupled oscillator system has two masses connected by springs. When analyzed, the system has natural frequencies $\omega_1 = \sqrt{2}$ rad/s and $\omega_2 = 2\sqrt{2}$ rad/s, with corresponding mode shapes $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$ . If the system starts from rest with initial displacements $x_1(0) = 3$ and $x_2(0) = 1$ , what is the displacement $x_1(t)$ ?

$x_1(t) = 2\cos(\sqrt{2}t) + \cos(2\sqrt{2}t)$ (correct answer)
$x_1(t) = 2\cos(\sqrt{2}t) - \cos(2\sqrt{2}t)$
$x_1(t) = \cos(\sqrt{2}t) + 2\cos(2\sqrt{2}t)$
$x_1(t) = 3\cos(\sqrt{2}t) + \cos(2\sqrt{2}t)$

Explanation: The general solution is

\mathbf{x}(t) = c_1\mathbf{v}_1\cos(\omega_1 t) + c_2\mathbf{v}_2\cos(\omega_2 t)

since the system starts from rest. At

t = 0

\mathbf{x}(0) = c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2\begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}

. This gives us

c_1 + c_2 = 3

and

c_1 - c_2 = 1

, so

c_1 = 2

and

c_2 = 1

. Therefore

x_1(t) = 2\cos(\sqrt{2}t) + 1\cos(2\sqrt{2}t) = 2\cos(\sqrt{2}t) + \cos(2\sqrt{2}t)

. Choice B incorrectly subtracts the second term. Choice C swaps the coefficients. Choice D uses the wrong coefficient for the first mode.

Question 3

Two masses $m_1 = 2$ kg and $m_2 = 1$ kg are coupled by springs such that the system matrix is $$A = \begin{pmatrix} -3 & 1 \ 2 & -4 \end{pmatrix}

One eigenvalue must be zero, but $\det(A) = 10 \neq 0$ , so no constraint exists
The trace of $A$ must equal the determinant, requiring $-7 = 10$
The system cannot be critically coupled with the given mass ratio
The determinant of $A$ must equal zero, requiring modification of spring constants (correct answer)

Explanation: When you encounter coupled mass-spring systems, the key insight is that critical coupling occurs when one natural frequency becomes zero, meaning the system has a zero eigenvalue. This happens when the system matrix becomes singular. For a system to have a zero eigenvalue, the determinant of the matrix

A

must equal zero. Let's check the current determinant:

\det(A) = (-3)(-4) - (1)(2) = 12 - 2 = 10

. Since this is non-zero, the system currently has no zero eigenvalues and cannot be critically coupled as given. To achieve critical coupling, you would need to modify the spring constants (which determine the entries of matrix

A

) so that

\det(A) = 0

. This makes answer D correct. Now for the wrong answers: A correctly identifies that one eigenvalue must be zero but incorrectly concludes that no constraint exists. The constraint is precisely that the determinant must be zero. B confuses the condition for critical coupling - there's no requirement that trace equals determinant. C is incorrect because critical coupling is achievable with any mass ratio by appropriately choosing spring constants. Remember this pattern: in coupled oscillator problems, critical coupling always corresponds to a zero eigenvalue, which requires a singular system matrix. When you see "critically coupled" or "critical damping" scenarios, immediately check whether the determinant equals zero - this is often the key constraint that determines the system parameters.

Question 4

Two pendulums of length $L$ and equal mass $m$ are coupled by a spring of constant $k$ attached at distance $d$ from their pivot points. For small oscillations, the linearized equations of motion are $mL^2\ddot{\theta_1} + mgL\theta_1 + kd^2(\theta_1 - \theta_2) = 0$ and $mL^2\ddot{\theta_2} + mgL\theta_2 + kd^2(\theta_2 - \theta_1) = 0$ . What is the frequency of the symmetric mode where both pendulums swing together?

$\omega_s = \sqrt{\frac{g}{L} + \frac{2kd^2}{mL^2}}$
$\omega_s = \sqrt{\frac{g}{L} + \frac{kd^2}{mL^2}}$
$\omega_s = \sqrt{\frac{g}{L}}$ (correct answer)
$\omega_s = \sqrt{\frac{g}{L} - \frac{kd^2}{mL^2}}$

Explanation: When analyzing coupled oscillator systems, the key insight is to look for normal modes—special motion patterns where the coupling terms simplify dramatically. This approach transforms a complex two-body problem into manageable single-oscillator equations. For the symmetric mode, both pendulums move identically:

\theta_1 = \theta_2

. This means they swing together as if connected by a rigid rod. Substituting this condition into either equation of motion, the coupling term

kd^2(\theta_1 - \theta_2)

becomes zero since

(\theta_1 - \theta_2) = 0

. The equation reduces to:

mL^2\ddot{\theta_1} + mgL\theta_1 = 0

, which gives

\ddot{\theta_1} + \frac{g}{L}\theta_1 = 0

. This is simple harmonic motion with frequency

\omega_s = \sqrt{\frac{g}{L}}

—exactly the frequency of a single uncoupled pendulum. Choice A incorrectly adds

\frac{2kd^2}{mL^2}

to the natural pendulum frequency, suggesting the spring somehow increases the restoring force even when both pendulums move together. Choice B adds

\frac{kd^2}{mL^2}

, making the same conceptual error with a different coefficient. Choice D subtracts the spring term, implying the coupling weakens the restoring force in symmetric motion—also incorrect since the spring doesn't stretch when both pendulums move identically. Study tip: In coupled oscillator problems, always check whether the proposed motion pattern eliminates coupling terms. Symmetric modes often behave like uncoupled systems because the coupling mechanism (here, the spring) experiences no deformation when both masses move identically.

Question 5

Consider a three-mass system where only masses 1 and 2 are coupled directly, but all three masses are connected to ground springs. If the coupling between masses 1 and 2 is removed, the individual frequencies are $\omega_1 = \sqrt{2}$ , $\omega_2 = \sqrt{3}$ , and $\omega_3 = \sqrt{5}$ rad/s. When weak coupling with strength $\epsilon$ is restored between masses 1 and 2, what is the first-order perturbation correction to the frequency of the middle mode?

$\Delta\omega_2 = \frac{\epsilon^2}{2\sqrt{3}(3-2)} = \frac{\epsilon^2}{2\sqrt{3}}$
$\Delta\omega_2 = 0$ because the middle mass is not directly affected by coupling (correct answer)
$\Delta\omega_2 = \frac{\epsilon}{\sqrt{3}}$ due to direct frequency shift
$\Delta\omega_2 = \frac{\epsilon^2}{\sqrt{3}}\left(\frac{1}{3-2} + \frac{1}{3-5}\right) = \frac{\epsilon^2}{\sqrt{3}}\left(1 - \frac{1}{2}\right)$

Explanation: In perturbation theory for coupled oscillators, if mass 2 is not directly coupled to masses 1 or 3, then the first-order correction to its frequency is zero because the coupling matrix element

\langle \phi_2^{(0)} | V | \phi_2^{(0)} \rangle = 0

where

V

represents the coupling between masses 1 and 2 only, and

\phi_2^{(0)}

is the unperturbed mode shape of mass 2. The coupling only affects mass 2's frequency through second-order corrections involving the other modes. Choice A gives a second-order correction formula. Choice C incorrectly assumes direct first-order coupling. Choice D attempts a second-order calculation but uses an incorrect formula for the case where mass 2 is not directly involved in the perturbing interaction.